are you sure that the inequality 0 < m (< n) is strict? Let p = q = 2. then for n = 2, we have that $2\cdot 2 = 2 + 2$, and so $pq = 2 \cdot 2 = 4 = n^2 - m^2 = 2^2 - m^2 = 4 - m^2$, so $m$ would need to be zero, else we have $n, p, q$ for which the antecedent is try, but the consequent false. Perhaps we need $0 \leq m < n$ with $m = 0 = n - p = 2 - 2, p = q$. However, I haven't even checked to verify that the "someone's" observation is true. Also, I suspect that we are to take m, n, p, q to non-negative integers?
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amWhyJun 29 '11 at 20:05

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I suspect that he overlooked the fact that one can have $n^2-m^2=pq$ for primes $p<q$ without having $p=n-m$ and $q=n+m$, since it may be that $n-m=1$, $n+m =pq$, and $2n=pq+1$.
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Brian M. ScottJun 29 '11 at 20:19

If you are able to read French, you might be interested in looking at the following link, in which I try to prove that the smallest $r$ such that $n-r$ and $n+r$ both are prime is such that $r=O(\log^2 n)$. If not, I'll try to give an English translation this weekend, tonight I feel too tired to do so. Meanwhile, anyone is welcome to give the desired translation if needed.